用csv在python中嵌套字典，我需要减少或增加日期

Question

我很兴奋，因为我从朋友那里得到了一些提示。我正在尝试使用循环创建字典字典。我正在使用的 csv 的日期为 2008 年至 2014 年，我使用日期作为键。 csv 看起来像这样：
年份，title_field，值

2014,Total Housing Units,49109
2014,Vacant Housing Units,2814
2014,Occupied Housing Units,46295
2013,Total Housing Units,47888
2013,Vacant Housing Units,4215
2013,Occupied Housing Units,43673
2012,Total Housing Units,45121
2012,Vacant Housing Units,3013
2012,Occupied Housing Units,42108
2011,Total Housing Units,44917
2011,Vacant Housing Units,4213
2011,Occupied Housing Units,40704
2010,Total Housing Units,44642
2010,Vacant Housing Units,3635
2010,Occupied Housing Units,41007
2009,Total Housing Units,39499
2009,Vacant Housing Units,3583
2009,Occupied Housing Units,35916
2008,Total Housing Units,41194
2008,Vacant Housing Units,4483
2008,Occupied Housing Units,36711

这是我的代码：

import csv

denton_housing = {}
filename = 'denton_housing.csv'
key = 2014

with open(filename, 'r', encoding='utf8', newline='') as f:
    for row in csv.DictReader(f, delimiter=','):
        while key not in denton_housing:
            denton_housing[key] = {}
            denton_housing[key][row['title_field']] = int(row['value'])
            key-1

当我打印出来时，我得到：

{2014: {'Total Housing Units': 49109}}

这太棒了！但我需要这个：

{2014: {'Total Housing Units': 49109}, {'Vacant Housing Units': 2814}, \    {'Occupied Housing Units': 46295}}

更重要的是，我需要它在 2013 年、2012 年、2011 年、2010 年、2009 年和 2008 年循环并执行相同的操作，但到此为止。

Answer 1

您可以使用itertools.groupby 创建一个字典，其中每个键是一年，使用Total Housing Units 和Vacant Housing Units 存储字典：

import itertools
import csv
with open('filename.csv') as f:
   data = [[int(a), b, int(c)] for a, b, c in sorted(list(csv.reader(f))[1:], key=lambda x:int(x[0]))]

final_data = {a:dict(i[1:] for i in b) for a, b in itertools.groupby(data, key=lambda x:x[0])}

输出：

{2008: {'Total Housing Units': 41194, 'Vacant Housing Units': 4483, 'Occupied Housing Units': 36711}, 2009: {'Total Housing Units': 39499, 'Vacant Housing Units': 3583, 'Occupied Housing Units': 35916}, 2010: {'Total Housing Units': 44642, 'Vacant Housing Units': 3635, 'Occupied Housing Units': 41007}, 2011: {'Total Housing Units': 44917, 'Vacant Housing Units': 4213, 'Occupied Housing Units': 40704}, 2012: {'Total Housing Units': 45121, 'Vacant Housing Units': 3013, 'Occupied Housing Units': 42108}, 2013: {'Total Housing Units': 47888, 'Vacant Housing Units': 4215, 'Occupied Housing Units': 43673}, 2014: {'Total Housing Units': 49109, 'Vacant Housing Units': 2814, 'Occupied Housing Units': 46295}}

Answer 2

In[2]: import csv
  ...: from collections import defaultdict
  ...: 
  ...: denton_housing = defaultdict(dict)
  ...: filename = 'denton_housing.csv'
  ...: key = 2014
  ...: 
  ...: with open(filename, 'r', encoding='utf8', newline='') as f:
  ...:     for row in csv.DictReader(f):
  ...:         denton_housing[row['year']].update({
  ...:             row['title_field']: int(row['value'])
  ...:         })
  ...: 
In[3]: import json
In[4]: print(json.dumps(denton_housing, indent=4))
{
    "2014": {
        "Total Housing Units": 49109,
        "Vacant Housing Units": 2814,
        "Occupied Housing Units": 46295
    },
    "2013": {
        "Total Housing Units": 47888,
        "Vacant Housing Units": 4215,
        "Occupied Housing Units": 43673
    },
    "2012": {
        "Total Housing Units": 45121,
        "Vacant Housing Units": 3013,
        "Occupied Housing Units": 42108
    },
    "2011": {
        "Total Housing Units": 44917,
        "Vacant Housing Units": 4213,
        "Occupied Housing Units": 40704
    },
    "2010": {
        "Total Housing Units": 44642,
        "Vacant Housing Units": 3635,
        "Occupied Housing Units": 41007
    },
    "2009": {
        "Total Housing Units": 39499,
        "Vacant Housing Units": 3583,
        "Occupied Housing Units": 35916
    },
    "2008": {
        "Total Housing Units": 41194,
        "Vacant Housing Units": 4483,
        "Occupied Housing Units": 36711
    }
}

Answer 3

这里的诀窍是，您不希望每个值中都有一个dict，而是需要dicts 中的list。 （实际上，如果您想要您指定的确切输出，tuple，但我认为您会对 list 感到满意。）

所以，不要这样：

denton_housing[key] = {}

……这样做：

denton_housing[key] = []

现在，而不是设置

而不是这个：

denton_housing[key][row['title_field']] = int(row['value'])

……这样做：

new_dict = {}
new_dict[row['title_field']] = int(row['value'])
denton_housing[key].append(new_dict)

另外，始终执行第二部分，而不仅仅是在key not in 时。 （另外，你可能想要if，而不是while。）

您可以从这里简化事情，并构建一个更好的数据结构（您有三个单键字典；一个具有多个键的字典不是更好吗？）等等。但希望这对你来说很容易理解，这样你就可以摆脱困境，从那里开始疯狂。